Well I know my question already has answers, but I was not able to follow them. Let us say that we have rotated the function y = $x^3$ around x-axis to obtain something like the image.
Now we want to find the surface area from x = 1 to x = 2.
The book says that the formula is:
$2\pi\int_1^2x^3\sqrt{1+9x^4}dx $
or in general, the surface area in these type of questions is:
$2\pi\int_a^bf(x)\sqrt{1+(f'(x))^2}dx $
I don't know why we need to take the length of the curve into account. The question requires splitting the solid into infinite cylinders with infinetesmially small width(dx). The radius of the circle is f(x). So, the curved surface area of 1 such cylinder must be $2\pi f(x)dx$. Therefore, the total surface area is:
$\int_a^b2\pi f(x)dx = 2\pi\int_a^bf(x)dx$
I don't know what is wrong with my approach or why should we take the arc length into consideration. Intuitively, my reasoning seems right.
Please answer in a simpler way as integrals have never been my strength.